$\begin{array}{1 1} 1.082 \times 10^6 \; gm^{-3} \\5.812\; g\;cm^{-3} \\ 6.324 \times 10^6 \; gm^{-3} \\ 4.346 \; g\;cm^{-3} \end{array}$

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Answer: $5.812\; g\;cm^{-3}$

Number of A ions per unit cell = $\large\frac{1}{8}$$\times 8 = 1$

Number of B ions per unit cell = $\large\frac{1}{2}$$\times 6 = 3$

$\Rightarrow$ the empirical formula is $AB_3$

$\Rightarrow$ the mass of the compound $ = 50 + 3 \times 100 = 350\; gm$

The density is given by the formula $\large\frac{z\;M}{a^3\;N_A}$

Avagadro's number $N_A = 6.022 \times 10^{23} /\; mol$

Given $z = 1,\; M = 350 \; gm, \; a = 10^{\circ}A = 10 \times 10^{-8}cm$, we can calculate density as follows:

$\Rightarrow$ Density $\rho = \large\frac{ 1 \times 350}{ (10 \times 10^{-8})^3 \times 6.022 \times 10^{23}}$$ = 5.812\; g\;cm^{-3}$

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